Question

Arata ca 1+4+7+...+301 supra 1+2+3+...+100=151 supra 50
URGENT VA ROG REZOLVARE SI RASPUNS
Dau coroana si 30 de puncte

Answer 1

1+4+7+...+301 = 1+1+3+1+3×2+1+3×3+...+1+3×100 = 1+1+...+1+3(1+2+3+...+100)
1+1+...+1 sunt 101 de numere 

$\frac{101+3(1+2+3+...+100)}{1+2+3+...+100} = \frac{151}{50}$

$\frac{101+3}{1} = \frac{151}{50}$

$104 = \frac{151}{50}$

Am gresit cumva undeva?